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Derivative of 1xx4\frac{\frac{1}{x}}{x^{4}}

The derivative of 1xx4\frac{\frac{1}{x}}{x^{4}} is −1x2x4−4x3x(x4)2\frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}}.

f(x) =
ddx[1xx4]=−1x2x4−4x3x(x4)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{4}}\right] = \frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}}

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Graph of f(x) = 1 / x / x^4 and its derivative-3-2-1123-40000-30000-20000-10000slope -5f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / x^4 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -5.

How to work it out

  1. 1

    We differentiate f(x)=1xx4f(x) = \frac{\frac{1}{x}}{x^{4}} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xx4f(x) = \frac{\frac{1}{x}}{x^{4}}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=1xu = \frac{1}{x} and v=x4v = x^{4}.

    u=1x,v=x4u = \frac{1}{x},\quad v = x^{4}
  3. 3

    Differentiate each piece:

    u′=−1x2,v′=4x3u' = \frac{-1}{x^{2}},\quad v' = 4 x^{3}
  4. 4

    Put them in the rule:

    f′(x)=(−1x2)(x4)−(1x)(4x3)(x4)2f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{4}\right) - \left(\frac{1}{x}\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=−1x2x4−4x3x(x4)2f'(x) = \frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}}
  6. 6

    Differentiating once more gives the second derivative, which tells you about curvature:

    f′′(x)=(−((−2)x)(x2)2x4+−1x24x3−12x2x−4x3x2)(x4)2−(−1x2x4−4x3x)2x44x3((x4)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{4} + \frac{-1}{x^{2}} 4 x^{3} - \frac{12 x^{2} x - 4 x^{3}}{x^{2}}\right) \left(x^{4}\right)^{2} - \left(\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{4}}\right] = \frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{x^{4}} \]
\[ u = \frac{1}{x},\quad v = x^{4} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = 4 x^{3} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{4}\right) - \left(\frac{1}{x}\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{4} + \frac{-1}{x^{2}} 4 x^{3} - \frac{12 x^{2} x - 4 x^{3}}{x^{2}}\right) \left(x^{4}\right)^{2} - \left(\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}} \]

Values of f and f' for 1 / x / x4x^{4}

xf(x)f'(x)
-2-0.03125-0.078125
-1-1-5
0.532-320
11-5
20.03125-0.078125
30.0041152263-0.0068587106

Questions

What is the derivative of 1 / x / x4x^{4}?

The derivative of 1xx4\frac{\frac{1}{x}}{x^{4}} is −1x2x4−4x3x(x4)2\frac{\frac{-1}{x^{2}} x^{4} - \frac{4 x^{3}}{x}}{\left(x^{4}\right)^{2}}.

What is the second derivative of 1 / x / x4x^{4}?

((-(-2x) / (x2x^{2})^2 * x4x^{4} + -1 / x2x^{2} * 4 * x3x^{3} - (12 * x2x^{2} * x - 4 * x3x^{3}) / x2x^{2}) * (x4x^{4})^2 - (-1 / x2x^{2} * x4x^{4} - 4 * x3x^{3} / x) * 2 * x4x^{4} * 4 * x3x^{3}) / ((x4x^{4})^2)^2.

Which rule do you use for 1 / x / x4x^{4}?

The quotient rule.

What is the slope of 1 / x / x4x^{4} at x = 1?

-5.

Related derivatives

    Derivative of 1 / x / x^4: (-1 / x^2 * x^4 - 4 * x^3 / x) / (x^4)^2 (with steps) | WordGPT